Question

Four different objects 1, 2, 3, 4 are distributed at random in four places marked 1,2, 3, 4. What is the probability that none of the objects occupy the place corresponding to their number?
A) $\dfrac{{17}}{{24}}$
B) $\dfrac{3}{8}$
C) $\dfrac{1}{2}$
D) $\dfrac{5}{8}$

Answer 1

We can find the total number of ways of arranging four objects in 4 places. Then we can find the number of ways of arranging the objects one in the corresponding place, 2 in the corresponding place and 3 in the corresponding place. Then we can find the number of ways of placing the objects such that none of the objects occupy the place corresponding to their number by finding their difference. Then we can find the required probability by dividing the number of ways of placing the objects such that none of the objects occupy the place corresponding to their number by the total possible arrangements.

Complete step by step solution:
We need to arrange 4 objects 1, 2, 3 ,4 in four places 1,2,3,4.
There are 4 objects to be placed on 4 places. So, it can be done in $4!$ ways.
$4! = 4 \times 3 \times 2 = 24$
So, the total number of ways of placing four objects is 24.
Now we can consider the case where 3 objects match with their places.
There is only one way where 3 objects match with their places. If we match any 3 places, then the 4th place also will be in the correct place.
$\Rightarrow {}^3{C_3} = 1$
Then we can consider the case where exactly two objects match. If we take any 2 then there is exactly one way for the other 2 don’t match. We can choose ${}^4{C_2}$such pairs.
As $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ , using this we get,
$\Rightarrow {}^4{C_2} = \dfrac{{4!}}{{2!\left( {4 - 2} \right)!}}$
On simplification we get,
$\Rightarrow {}^4{C_2} = \dfrac{{4 \times 3 \times 2}}{{2 \times 2}}$
On further simplification we get,
${ \Rightarrow ^4}{C_2} = 6$
Now consider the case where only one number matches. If a take the number 1 to be in its position, there are only 2 possible arrangement which make the other 3 numbers in different places. We have 4 such numbers, so there are $4 \times 2 = 8$ ways

So, the number of ways no object occupies the corresponding place is given by the subtracting the sum of all the 3 cases from the total possible ways.
$\Rightarrow 24 - \left( {8 + 6 + 1} \right) = 24 - 15$
=9
So, the required probability is given by dividing the number of ways of placing the objects such that none of the objects occupy the place corresponding to their number by the total possible arrangements.
$\Rightarrow P = \dfrac{9}{{24}}$
On simplification we get,
$\Rightarrow P = \dfrac{3}{8}$
Therefore, the required probability is $\dfrac{3}{8}$

So, the correct answer is option C which is $\dfrac{3}{8}$.

Note:
Note: Alternate method to solve this problem is by,
We can list all the possible ways of arrangements
$\left( {1,2,3,4} \right)\left( {1,2,4,3} \right)\left( {1,3,2,4} \right)\left( {1,4,3,2} \right)\left( {1,4,2,3} \right)\left( {1,3,4,2} \right)$
$\left( {2,1,3,4} \right)\left( {2,1,4,3} \right)\left( {2,3,1,4} \right)\left( {2,3,4,1} \right)\left( {2,4,1,3} \right)\left( {2,4,3,1} \right)$
$\left( {3,1,2,4} \right)\left( {3,1,4,2} \right)\left( {3,2,1,4} \right)\left( {3,2,4,1} \right)\left( {3,4,1,2} \right)\left( {3,4,2,1} \right)$
$\left( {4,1,2,3} \right)\left( {4,1,3,2} \right)\left( {4,2,1,3} \right)\left( {4,2,3,1} \right)\left( {4,3,2,1} \right)\left( {4,3,1,2} \right)$
There are 24 possible arrangements, out of which only 9 of the following are the favourable ones.
$\left( {2,1,4,3} \right)\left( {2,3,4,1} \right)\left( {2,4,1,3} \right)\left( {3,1,4,2} \right)\left( {3,4,1,2} \right)\left( {3,4,2,1} \right)\left( {4,1,2,3} \right)\left( {4,3,2,1} \right)\left( {4,3,1,2} \right)$
So, the required probability is given by the number of favourable outcomes divided by total outcomes.
$\Rightarrow P = \dfrac{9}{{24}}$
On simplification we get,
$\Rightarrow P = \dfrac{3}{8}$
Therefore, the required probability is $\dfrac{3}{8}$.